How to go from sum of products to product of sums?

Multiplication gives us distribution over the products, so

(a′+b+d′) (a′+b+c′+f′) = a′ (a′+b+c′+f′) + b (a′+b+c′+f′) + d′ (a′+b+c′+f′)

And then you can then distribute again each of the factors on the right.

Then you should simplify in any given number of ways. To take as an example, you have a′b and ba′, and since a′b + a′b = a′b + a′b = a′b, you can just drop one of them. Since bb = b, you can rewrite bb as b and etc.

So in the end part we should arrive at a sum of products. Then you can just invert. For example, if at the end you had:

p′ = a′b + bc′ + d′f ′+ a′f′

Then we would have

p = p′′ = (a′b + bc′ + d′f′ + a′f′)′ = (a′b)′⋅(bc′)′⋅(d′f′)′⋅(a′f′)′

Then applying De Morgan's laws to each of the factors, e.g., (a′b)′ = a+b′, so we would have

p = (a+b′)⋅(b′+c)⋅(d+f)⋅(a+f)

which is a product of sums.

54

step-by-step explanation:

6x9

well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.